Problem: Let $g(x)=\sin(4x^2+3x)$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(8x+3)\cos(4x^2+3x)$ (Choice B) B $\cos(4x^2+3x)$ (Choice C) C $(8x+3)\sin(4x^2+3x)$ (Choice D) D $\cos(8x+3)$
$g$ is a trigonometric function, but its argument isn't simply $x$. Therefore, it's a composite trigonometric function. In other words, suppose $u(x)=4x^2+3x$, then $g(x)=\sin\Bigl(u(x)\Bigr)$. $g'(x)$ can be found using the following identity: $\dfrac{d}{dx}\left[\sin\Bigl(u(x)\Bigr)\right]={\cos\Bigl(u(x)\Bigr)\cdot u'(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\sin(4x^2+3x) \\\\ &=\dfrac{d}{dx}\sin\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=4x^2+3x} \\\\ &={\cos\Bigl(u(x)\Bigr)\cdot u'(x)} \\\\ &=\cos\left(4x^2+3x\right)\cdot(8x+3)&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=(8x+3)\cos\left(4x^2+3x\right) \end{aligned}$ In conclusion, $g'(x)=(8x+3)\cos\left(4x^2+3x\right)$.